Hartshorne IV.6.4 - no curve of degree 9 and genus 11 in P^3 I'm working on this exercise in Hartshorne: there are no curves of degree 9 and genus 11 in $\mathbb{P}^3$. 
The hint says to show that it would have to lie on a quadric surface. This is the part I'm having trouble with. If $X$ is the curve, then there's an exact sequence
$$0 \to H^0(\mathcal{I}_X(2)) \to H^0(\mathcal{O}_{\mathbb{P}^3}(2)) \to H^0(\mathcal{O}_X(2)) \to \ldots$$
and I want $h^0(\mathcal{I}_X(2))$ to be nonzero, which would be true if $h^0(\mathcal{O}_X(2)) < h^0(\mathcal{O}_{\mathbb{P}^3}(2)) = 10$. If $D = \mathcal{O}_X(2)$, then Riemann-Roch says
$$h^0(D) - h^0(K - D) = \deg D + 1 - g = 2 \cdot 9 + 1 - 11 = 8$$
so if $h^0(K - D) = 0$, then all is good. But if $h^0(K - D) \ne 0$, then I can only think of Clifford's theorem, which says
$$h^0(D) - 1 \le \frac{1}{2} \deg D = 9$$
with equality iff $D = 0, K$, or $X$ is hyperelliptic and $D$ is a multiple of the $g^1_2$. If the inequality is strict, then again all is good, so this just leaves the case $X$ hyperelliptic, $D = 9 \cdot g^1_2$. Why is this impossible? Is there a simpler argument altogether?
 A: First, we show that any such curve $X$ must lie on a quadratic surface. Consider the sequence$$0 \to \mathcal{I}_X(2) \to \mathcal{O}_{\mathbb{P}^3}(2) \to \mathcal{O}_X(2) \to 0.$$We have $\text{dim}\,H^0(\mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(2)) = 10$, so if $\text{dim}\,H^0(X, \mathcal{O}_X(2)) < 10$, then $\text{dim}\,H^0(\mathbb{P}^3, \mathcal{I}_X(2)) \ge 1$. Thus, $X$ lies on a quadratic surface. $\mathcal{O}_X(2)$ corresponds to degree $2$ hypersurfaces, which will intersect the curve $X$ in $9(2) = 18$ points, so $\text{deg}\,|\mathcal{O}_X(2)| = 18$. For $\mathcal{O}_X(2)$ nonspecial, Riemann-Roch gives$$\text{dim}\,H^0(X, \mathcal{O}_X(2)) = 18 + 1 - 11 = 9,$$so $X$ will lie on a quadratic hypersurface. For $\mathcal{O}_X(2)$ special, and effective divisor $D$ in the linear system given by $\mathcal{O}_X(2)$ must have by Clifford's theorem$$\dim|D| \le {1\over2}\text{deg}\,D = 9,$$so again $X$ will lie on a quadratic hypersurface.
Now, suppose $X$ lies on a nonsingular quadratic hypersurface of type $(a, b)$. Then by Remark 6.4.1, $d = 9 = a + b$ and $g = 11 = ab - a - b + 1$. Substituting, $11 = a(9 - a) - a - (9 - a) + 1$, or $0 = a^2 - 9a + 19$, which has no integer solution. Thus, $X$ can not lie on a nonsingular quadratic hypersurface.
$X$ can not lie in the product of two hyperplanes, since it will then either be a line and have genus $0$, or it will be in a plane, and then contradicts$$g = {1\over2}(d - 1)(d - 2) = 28 \neq 11.$$The only case left is $X$ lying on a quadratic cone, but then Remark 6.4.1 again gives$$d = 2a + 1 \implies a = 4,$$and then$$g = a^2 - a \implies g = 16 - 4 = 12 \neq 11.$$Thus, no curve exist in $\mathbb{P}^3$ of degree $9$ and genus $11$.
A: You just need to exclude the case when $dimH^{0}(X,\mathcal{O}_{X}(2))=10$, $X$ is hyperelliptic and $D=9g_{2}^{1}\in |\mathcal{O}_{X}(2)|$.
If this is the case, by Rieman-Roch theorem, we have
$$H^{0}(X,\mathcal{O}_{X}(D))-H^{0}(X,\mathcal{O}_{X}(K_{X}-D))=8$$
Since $H^{0}(X,\mathcal{O}_{X}(D))=10$, we have $H^{0}(X,\mathcal{O}_{X}(K_{X}-D))=2$. $deg|K_{X}-D|=2g-2-deg(D)=2$. We have $|K_{X}-D|$ is a linear system of degree $2$ dimension $1$. $K_{X}-D=g_{2}^{1}$. We have $$K_{X}=D+g_{2}^{1}$$.
Since $D=2H$ is very ample and $g_{2}^{1}$ is ample, we conclude $K_{X}=D+g_{2}^{1}$ is very ample, which contradicts to our assumption that $X$ is hyperellptic.
